The ratio of the radius of the first Bohr orbit to that of the second Bohr orbit of the orbital electron is

$A$ hydrogen atom changes its state from $n=3$ to $n=2$. Due to recoil,the percentage change in the wavelength of emitted light is approximately $1 \times 10^{-n}$. The value of $n$ is . . . . . . .[Given $Rhc=13.6 \text{ eV}, hc=1242 \text{ eV nm}, h=6.6 \times 10^{-34} \text{ J s}$,mass of the hydrogen atom $=1.67 \times 10^{-27} \text{ kg}$]

The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. The potential and kinetic energies of the electron in this state are . . . . . . .

In a hydrogen atom,an electron makes a transition from the $n=4$ to the $n=1$ energy level. The recoil momentum of the $H$ atom will be:

The following parameter is the same for all hydrogen-like atoms and ions in their ground state.

Assume that there is no repulsive force between the electrons in an atom,but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstances,calculate the ground state energy of a $He$ atom.